Chapter 7, Problem 47E

a.

Explanation of Solution

Solution:

“No”, database involves 80% of the system activities which might get affected if the disk access time gets doubled.

$\begin{array}{l}\text{Percentage of transaction on the disk × half of the throughput}\\ \text{= 0}\text{.8 × }\frac{\text{10000}}{\text{2}}\\ \text{= \$4000}\end{array}$

$\begin{array}{}\end{array}$

b.

Explanation of Solution

Solution:

Like the read operation in RAID-1, the write operation doubles the time as read because of its mirroring capability. If the system doesn’t use RAID-1, the access time for read operation will be half of the value. Consider that the disk arms are 180 degrees offset from one another.

$\begin{array}{l}\text{Average access time = [0}\end{array}$

c.

Explanation of Solution

Solution:

For the disk to be free, then 25% of the transaction must wait behind one transaction. If the system uses RAID-5 along with 8 disks (two sets of 4 disks) then the average disk access time would be:

$\begin{array}{l}\text{Average access time = [0}\text{.75 × 15ms]+ [0}\end{array}$

d.

Explanation of Solution

Solution:

RAID-1:

• It requires 2*N disks.
• The cost of this will be $24,000.

2 sets of 4 disk RAID-5:

• It requires 133% of the number of disks.
• The cost of this will be $16,000.
• The MTTF is given as 20,000 hours...